If 0.582 moles of zinc reacts with excess lead(iv) sulfate how many grams are zinc sulfate would be produced?

94.0 g

Explanation:

1) Reactants:

i) zinc (limiting)

ii) lead(IV) sulfate (excess); Pb(SO₄)₂

2) Chemical equation:

2 Zn + Pb(SO₄)₂ → 2 ZnSO4 + Pb

3) mole ratios

2 mol Zn: 1 mol Pb(SO₄)₂ : 2 mole ZnSO₄ : 1 mol Pb

4) Proportion

x / 0.582 mol Zn = 2 mole ZnSO₄ / 2 mol Zn

⇒ x = 1 × 0.582 mole ZnSO₄ = 0.582 mol ZnSO₄

5) Convert 0.582 mole ZnSO₄ to grams

mass in grams = number of moles × molar mass

molar mass ZnSO₄ = [65.39 + 32.065 + 4 × 15.999] g/mol = 161.451 g/mol

mass in grams = 0.582 mol × 161.451 g/mol = 93.96 g

Round to three significant figures: 94.0 g

since the atomic number is the number of protons in the nucleus, as we move to the left on the periodic table, the electrostatic pull between the nucleus and the valence electron shell decreases. as we move down a group (vertical column), the valence shell energy level gets further and further from the nucleus.